Problem: Solve for $x$, ignoring any extraneous solutions: $\dfrac{x^2 - 8x}{x - 3} = \dfrac{4x - 27}{x - 3}$
Multiply both sides by $x - 3$ $ \dfrac{x^2 - 8x}{x - 3} (x - 3) = \dfrac{4x - 27}{x - 3} (x - 3)$ $ x^2 - 8x = 4x - 27$ Subtract $4x - 27$ from both sides: $ x^2 - 8x - (4x - 27) = 4x - 27 - (4x - 27)$ $ x^2 - 8x - 4x + 27 = 0$ $ x^2 - 12x + 27 = 0$ Factor the expression: $ (x - 9)(x - 3) = 0$ Therefore $x = 9$ or $x = 3$ However, the original expression is undefined when $x = 3$. Therefore, the only solution is $x = 9$.